American Options

American OptionsThis and the following chapters form part of the course Stochastic Differential Equations for Fi-nance II.. Suppose we exercise the first time the stock price isLor lowe

American Options

This and the following chapters form part of the course Stochastic Differential Equations for

Fi-nance II.

dS = rS dt + S dB Intrinsic value at timet : ( K,S ( t )) + :

LetL2[0 ;K ]be given Suppose we exercise the first time the stock price isLor lower We define

 L = minft0; S ( t )Lg;

v L( x ) = IEe,rL( K,S (  L)) +

=

(

K,x ifxL,

( K,L ) IEe,rL ifx > L:

The plan is to comutev L( x )and then maximize overLto find the optimal exercise price We need

to know the distribution of L

(Based on the reflection principle)

LetBbe a Brownian motion underIP, letx > 0be given, and define

 = minft0; B ( t ) = xg:

 is called the first passage time tox We compute the distribution of

247

K

x

Stock price

Figure 25.1: Intrinsic value of perpetual American put

Define

M ( t ) = max 0

ut B ( u ) : From the first section of Chapter 20 we have

IPfM ( t )2dm;B ( t )2dbg= 2(2 m,b )

tp

2 t exp

( ,

(2 m,b ) 2

2 t

)

dm db; m > 0 ;b < m: Therefore,

IPfM ( t ) xg=Z

1

x

Z m

,1

2(2 m,b )

tp

2 t exp

( ,

(2 m,b ) 2

2 t

)

db dm

=Z 1

p

2 t exp

( ,

(2 m,b ) 2

2 t

)

b=m b=,1

dm

=Z 1

p

2 t exp

( ,

m 2

2 t

) dm:

We make the change of variablez = pm

tin the integral to get

=Z 1

x=p

t 2

p

2  exp

( ,

z 2 2

) dz:

Now

 t( )M ( t )x;

IPf 2dtg= @

@tIPf tgdt

= @

@tIPfM ( t )xg dt

=

"

@

@t

Z 1

x=p

t 2

p

2  exp

( ,

z 2 2

) dz

# dt

=,

2

p

2  exp

( ,

x 2

2 t

) : @@t

x p t

 dt

tp

2 t exp

( ,

x 2

2 t

) dt:

We also have the Laplace transform formula

IEe, =Z

1

0 e,t IPf 2dtg

= e,xp

2 ; > 0 : (See Homework) Reference: Karatzas and Shreve, Brownian Motion and Stochastic Calculus, pp 95-96

Reference: Karatzas/Shreve, Brownian motion and Stochastic Calculus, pp 196–197.

For0t <1, define

e

B ( t ) = t + B ( t ) ;

Z ( t ) = expf,B ( t ),1 2  2 tg;

= expf, Be( t ) + 1 2  2 tg; Define

~

 = minft0; Be( t ) = xg:

We fix a finite timeT and change the probability measure “only up toT” More specifically, with

T fixed, define

f

IP ( A ) =Z

A Z ( T ) dP; A2 F( T ) : UnderfIP, the processBe( t ) ; 0tT, is a (nondrifted) Brownian motion, so

f

IPf ~2dtg= IPf 2dtg

tp

2 t exp

( ,

x 2

2 t ) dt; 0 < tT:

For0 < tT we have

IPf ~tg= IEh

1f ~tg i

= IEf 

1f ~tg

1

Z ( T )



= IEf h

1f ~tgexpf Be( T ),1 2  2 Tg

i

= IEf 

1f ~tg f

IE

expf Be( T ),1 2  2 Tg

F(~  ^t )

= IEf h

1f ~tgexpf Be(~  ^t ),1 2  2 (~  ^t )g

i

= IEf h

1f ~tgexpfx,1 2  2  ~g

i

=Z t

0 expfx,1 2  2 sg

f

IPf ~2dsg

=Z t

sp

2 s exp

(

x,

1

2  2 s,

x 2

2 s

) ds

=Z t

sp

2 s exp

( ,

( x,s ) 2

2 s

) ds:

Therefore,

IPf ~2dtg= x

tp

2 t exp

( ,

( x,t ) 2

2 t

) dt; 0 < tT:

SinceT is arbitrary, this must in fact be the correct formula for allt > 0

Recall the Laplace transform formula for

 = minft0; B ( t ) = xg for nondrifted Brownian motion:

IEe, =Z

1

tp

2 t exp

( ,t,

x 2

2 t

)

dt = e,xp

2 ; > 0 ;x > 0 :

For

~

 = min t 0; t + B ( t ) = x ;

the Laplace transform is

IEe,~ =Z

1

tp

2 t exp

( ,t,

( x,t ) 2

2 t

) dt

=Z 1

tp

2 t exp

( ,t,

x 2

2 t + x,1 2  2 t

) dt

= e xZ

1

tp

2 t exp

( ,( + 1 2  2 ) t,

x 2

2 t

) dt

= e x,xp

2+2

; > 0 ;x > 0 ; where in the last step we have used the formula forIEe, withreplaced by + 1 2  2.

If ~ ( ! ) <1, then

lim #0 e,~( ! ) = 1;

if ~ ( ! ) =1, thene,~( ! ) = 0for every > 0, so

lim #0 e,~( ! ) = 0 : Therefore,

lim #0 e,~( ! ) = 1 < ~ 1: Letting#0and using the Monotone Convergence Theorem in the Laplace transform formula

IEe,~ = e x,xp

2+2

;

we obtain

IPf < ~ 1g= e x,xp

2

= e x,xjj:

If0, then

IPf < ~ 1g= 1 :

If < 0, then

IPf~  <1g= e 2x < 1 : (Recall thatx > 0)

(Based on martingales)

Let > 0be given Then

Y ( t ) = expfB ( t ),

1

2  2 tg

is a martingale, soY ( t^ )is also a martingale We have

1 = Y (0^ )

= IEY ( t^ )

= IE expfB ( t^ ),1 2  2 ( t^ )g:

= lim t

!1

IE expfB ( t^ ),1 2  2 ( t^ )g:

We want to take the limit inside the expectation Since

0  expfB ( t^ ),

1

2  2 ( t^ )g  e x ; this is justified by the Bounded Convergence Theorem Therefore,

1 = IE t lim

!1

expfB ( t^ ),

1

2  2 ( t^ )g: There are two possibilities For those!for which ( ! ) <1,

lim

t!1

expfB ( t^ ),1 2  2 ( t^ )g= e x,1 2 2 : For those!for which ( ! ) =1,

lim

t!1

expfB ( t^ ),1 2  2 ( t^ )g t lim

!1

expfx, 1 2  2 tg = 0 : Therefore,

1 = IE t lim

!1

expfB ( t^ ),1 2  2 ( t^ )g

= IE e x,1 2 2 1 <1



= IEe x,1 2 2 ; where we understande x,1 2 2 to be zero if =1

Let = 1 2  2, so =p

2 We have again derived the Laplace transform formula

e,xp

2 = IEe, ; > 0 ;x > 0 ; for the first passage time for nondrifted Brownian motion

dS = rS dt + S dB

S (0) = x

S ( t ) = x expf( r,1 2  2 ) t + B ( t )g

= x exp

8

>

<

>

:



2 6 6 4

 r

 ,



2



| {z }



t + B ( t )

3 7 7 5

9

>

=

>

; :

Intrinsic value of the put at timet:( K,S ( t )) +.

LetL2[0 ;K ]be given Define forxL,

 L = minft0; S ( t ) = Lg

= minft0; t + B ( t ) = 1  log L xg

= minft0; ,t,B ( t ) = 1  log L xg Define

v L = ( K,L ) IEe,rL

= ( K,L )exp

,



 log L x ,

1

 log L x

p

2 r +  2

= ( K,L )

x L

 ,, 1

p

2r+2 :

We compute the exponent

,



 ,

1



p

2 r +  2 =,

r

 2 + 1 2 ,

1



s

2 r +

r

 ,= 22

=,

r

 2 + 1 2 ,

1



s

2 r + r 2

 2 ,r +  2 = 4

=,

r

 2 + 1

2 ,

1



s

r 2

 2 + r +  2 = 4

=,

r

 2 + 1 2 ,

1



s

 r

 + = 2

2

=,

r

 2 + 1 2 ,

1



 r

 + = 2



=,

2 r

 2 : Therefore,

v L( x ) =

8

<

:

( K,L ),

xL
,2r= 2

; xL:

The curves( K,L ),

xL
,2r= 2

;are all of the formCx,2r= 2

We want to choose the largest possible constant The constant is

C = ( K L ) L 2r= 2

;

-2r/

K

K

x Stock price

K - x

(K - L) (x/L)

Figure 25.2: Value of perpetual American put

σ 2

-2r/

C1 x

σ 2

-2r/

x C2

σ 2

-2r/

x C3

x Stock price

Figure 25.3: Curves.

@C

@L =,L 2r2 + 2 r

 2 ( K,L ) L 2r2

,1

= L 2r2

 ,1 + 2 r

 2 ( K,L ) 1 L



= L 2r2

 ,



1 + 2 r

 2



+ 2 r

 2 K L

 :

We solve

,



1 + 2 r

 2



+ 2 r

 2 K

L = 0

to get

L = 2 rK

 2 + 2 r:

Since0 < 2 r <  2 + 2 r;we have

0 < L < K:

Solution to the perpetual American put pricing problem (see Fig 25.4):

v ( x ) =

8

<

:

( K,x ) ; 0xL;

( K,L),

xL

,2r= 2

; xL; where

L

= 2 rK

 2 + 2 r:

Note that

v0( x ) =

(

, 2r2( K,L )( L) 2r= 2

x,2r= 2

,1 ; x > L:

We have

lim

x#L

v0( x ) =,2 r

 2 ( K,L) 1 L

=,2 r

 2



K,

2 rK

 2 + 2 r



 2 + 2 r

2 rK

=,2 r

 2  2 + 2 r,2 r

 2 + 2 r

!

 2 + 2 r

2 r

=,1

= lim x

"L

v0( x ) :

-2r/

K

K

x Stock price

K - x

(K - L )(x/L ) * *

* L

Figure 25.4: Solution to perpetual American put.

Set

= 2 r

 2 ; L= 2 rK

 2 + 2 r = + 1 K:

If0x < L

, thenv ( x ) = K,x IfL

x <1, then

v ( x ) = ( K,L)( L)

(7.1)

= IE xh

e,r ( K,L) + 1f<1g

i

where

 = minft0; S ( t ) = L

If0x < L

, then

,rv ( x ) + rxv0( x ) + 1 2  2 x 2 v00( x ) = ,r ( K,x ) + rx (,1) = ,rK:

IfL

x <1, then

,rv ( x ) + rxv0( x ) + 1 2  2 x 2 v0( x )

= C [,rx,

,

, ,1

,1 2  2 x 2 (, ,1) x, ,2 ]

= Cx, [,r, ,1 2  2 (, ,1)]

= C (, ,1) x,



r,1 2  2

2 r

 2



= 0 :

In other words,vsolves the linear complementarity problem: (See Fig 25.5).

6

v K

K

L

@

@

@

@

@

@

@

@

@

@

Figure 25.5: Linear complementarity

For allx2IR,x6= L

,

rv,rxv0

,1 2  2 x 2 v0

v( K,x ) + ; (b) One of the inequalities (a) or (b) is an equality (c) The half-line[0 ;1)is divided into two regions:

C=fx ; v ( x ) > ( K,x ) +g;

S=fx ; rv,rxv0

,1 2  2 x 2 v0 > 0g; andL

is the boundary between them If the stock price is inC, the owner of the put should not exercise (should “continue”) If the stock price is inS or atL

, the owner of the put should exercise (should “stop”)

LetS (0)be given Sell the put at time zero forv ( S (0)) Invest the money, holding
( t )shares of stock and consuming at rateC ( t )at timet The valueX ( t )of this portfolio is governed by

dX ( t ) =
( t ) dS ( t ) + r ( X ( t ),
( t ) S ( t )) dt , C ( t ) dt;

or equivalently,

d ( e,rt X ( t )) =,e,rt C ( t ) dt + e,rt
( t ) S ( t ) dB ( t ) :

The discounted value of the put satisfies

d

e,rt v ( S ( t ))

= e,rth

,rv ( S ( t )) + rS ( t ) v0( S ( t ))+ 1 2  2 S 2 ( t ) v0( S ( t ))i

dt

+ e,rt S ( t ) v0

( S ( t )) dB ( t )

=,rKe,rt 1fS(t)<L dt + e,rt S ( t ) v0

( S ( t )) dB ( t ) :

We should set

C ( t ) = rK 1fS(t)<L ;

( t ) = v0

( S ( t )) :

, then

v ( S ( t )) = K,S ( t ) ;
( t ) = v0( S ( t )) =,1 :

To hedge the put whenS ( t ) < L

, short one share of stock and holdK in the money market As long as the owner does not exercise, you can consume the interest from the money market position, i.e.,

C ( t ) = rK 1fS(t)<L : Properties ofe,rt v ( S ( t )):

1 e,rt v ( S ( t ))is a supermartingale (see its differential above)

2 e,rt v ( S ( t ))e,rt ( K,S ( t )) +, 0t <1;

3 e,rt v ( S ( t ))is the smallest process with properties 1 and 2

Y ( t )e,rt ( K,S ( t )) + ; 0t <1: (8.1) Then property 3 says that

Y ( t ) e,rt v ( S ( t )) ; 0t <1: (8.2)

We use (8.1) to prove (8.2) fort = 0, i.e.,

Iftis not zero, we can taket to be the initial time andS ( t ) to be the initial stock price, and then adapt the argument below to prove property (8.2)

Y (0) 

|{z}

(8:1) ( K,S (0)) + = v ( S (0)) :

Case II:S (0) > L

: ForT > 0, we have

Y (0)IEY (  ^T ) (Stopped supermartingale is a supermartingale)

IEh

Y ( ^T ) 1f<1g

i : (SinceY 0) Now letT! 1to get

Y (0)T lim

!1

IEh

Y ( ^T ) 1f<1g

i

IEh

Y (  ) 1f<1g

i (Fatou’s Lemma)

IE 2 6

4e,r ( K,S (  )

| {z }

L

) + 1f<1g

3 7

5 (by 8.1)

= v ( S (0)) : (See eq 7.2)

Intinsic value:h ( S ( t ))

Value of the American contingent claim:

v ( x ) = sup  IE x

e,r h ( S (  ))

; where the supremum is over all stopping times

Optimal exercise rule: Any stopping time which attains the supremum

1 e,rt v ( S ( t ))is a supermartingale;

2 e,rt v ( S ( t ))e,rt h ( S ( t )) ; 0 < t <1;

3 e,rt v ( S ( t ))is the smallest process with properties 1 and 2

v ( x ) = sup  IE x

e,r ( S (  ),K ) +

Theorem 10.63

v ( x ) = x x 0 :

Proof: For everyt,

v ( x )IE xh

e,rt ( S ( t ),K ) +i

IE xh

e,rt ( S ( t ),K )i

= IE xh

e,rt S ( t )i

,e,rt K

= x,e,rt K:

Lett! 1to getv ( x )x

Now start withS (0) = xand define

Y ( t ) = e,rt S ( t ) : Then:

1 Y is a supermartingale (in fact,Y is a martingale);

2 Y ( t )e,rt ( S ( t ),K ) + ; 0t <1

Therefore,Y (0)v ( S (0)), i.e.,

xv ( x ) :

IE x

e,r ( S (  ),K ) +

< IE x

e,r S (  )

 x = v ( x ) : There is no optimal exercise time

Expiration time:T > 0

Intrinsic value:( K,S ( t )) +.

Value of the put:

v ( t;x ) =(value of the put at timetifS ( t ) = x)

= t sup

T

| {z }

:stopping time

IE x e,r(,t) ( K,S (  )) + :

See Fig 25.6 It can be shown thatv;v t ;v x are continuous across the boundary, whilev xx has a jump

LetS (0)be given Then

-x

t

L

T

K

v > K,x

,rv + v t + rxv x + 1 2  2 x 2 v xx = 0 v ( T;x ) = 0 ; xK

v = K,x

v t = 0 ; v x =,1 ; v xx = 0

,rv + v t + rxv x + 1 2  2 x 2 v xx =,rK

v ( T;x ) = K,x; 0xK

Figure 25.6: Value of put with expiration

1 e,rt v ( t;S ( t )) ; 0tT;is a supermartingale;

2 e,rt v ( t;S ( t ))e,rt ( K,S ( t )) + ; 0tT;

3 e,rt v ( t;S ( t ))is the smallest process with properties 1 and 2

Expiration time:T > 0

Intrinsic value:h ( S ( t ))

Value of the contingent claim:

v ( t;x ) = sup t

T IE x e,r(,t) h ( S (  )) : Then

rv,v t,rxv x, 1 2  2 x 2 v xx 0 ; (a)

vh ( x ) ; (b)

At every point( t;x )2[0 ;T ][0 ;1), either (a) or (b) is an equality (c)

1 e,rt v ( t;S ( t )) ; 0tT;is a supermartingale;

2 e,rt v ( t;S ( t ))e,rt h ( S ( t ));

3 e,rt v ( t;S ( t ))is the smallest process with properties 1 and 2

The optimal exercise time is

 = minft0; v ( t;S ( t )) = h ( S ( t ))g

If ( ! ) =1, then there is no optimal exercise time along the particular path!

K - x

(K - L) (x/L)

Figure 25.2: Value of perpetual American put

σ 2

-2r/

C1 x... < r <  + r;we have

0 < L < K:

Solution to the perpetual American put pricing problem (see Fig 25.4):

v ( x ) =

8

<... * *

* L

Figure 25.4: Solution to perpetual American put.

Set

= r

 ; L= rK